Solution to if random variable is uniformly distributed (-2,1) find pdf of y = 2x ^3 - Sikademy
Author Image

Archangel Macsika

if random variable is uniformly distributed (-2,1) find pdf of y = 2x ^3

The Answer to the Question
is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

Get the Answers Now!

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question

Since the random variable X is uniformly distributed with the interval, (-2,1), its pdf is given as,

f(x)={1\over b-a}={1\over 1--2}={1\over3}  

Therefore, the probability density function for the random variable X IS,

f(x)={1\over3},\space -2\leq x\leq 1\\0,\space elsewhere

To determine the pdf of y=2x^3, we shall apply the cumulative density function (CDF) method as described below.

 We determine G(y), the  cdf of y=2x^3,

Now,

G(y)=p(Y\leq y)=p(2x^3\leq y)\implies p(x\leq({y\over2})^{1\over3})

From definition of probability for continuous distributions,

 p(x\leq({y\over2})^{1\over3})=\displaystyle\int^{({y\over2})^{1\over3}}_{-2}f(x)dx

So,

G(y)=p(x\leq({y\over2})^{1\over3})=\displaystyle\int^{({y\over2})^{1\over3}}_{-2}{1\over3}dx={x\over3}|^ {({y\over2})^{1\over3}} _{-2}= {{({y\over2})^{1\over3}}\over3}+{2\over3}

To find the pdf of the random variable y, we differentiate G(y) with respect to y. That is,

g(y)={dG(y)\over dy}={{1\over3}\times{1\over2}\times({y\over2})^{-{2\over3}}\over3}={1\over18}({y\over2})^{-{2\over3}}

The limits are,

Lower limit,

y=2\times(-2)^3=-16

Upper limit,

y=2\times(1)^3=2

Therefore, the probability density function of the random variable y  is,

 g(y)={1\over18}({y\over2})^{-{2\over3}},\space -16\leq y\leq 2\\0,\space elsewhere


Related Answers

Was this answer helpful?

Join our Community to stay in the know

Get updates for similar and other helpful Answers

Question ID: mtid-4-stid-46-sqid-2517-qpid-987